The velocity - time graph of a ball of mass 25g moving on road is as given below:
1 answer:
Assume that the ball rolls to the right.
From the velocity-time graph, the acceleration is
a = (0 - 10 m/s)/(4 - 0 s) = -2.5 m/²
Part (a)
Because the mass of the ball is 25g or 0.025 kg, the force that the road exerts on the ball is
F = (0.025 kg)*(-2.5 m/s²) = -0.0625 N
The force is negative because it acts in opposition to the motion of the ball.
Answer: 0.0625 N to the left.
Part (b)
The direction of the force exerted by the road is to the left because it opposes the motion of the ball to the right.
Part (c)
The unit force is 1 N (Newton), with the dimension (kg-m)/s².
From the velocity-time graph, the acceleration is
a = (0 - 10 m/s)/(4 - 0 s) = -2.5 m/²
Part (a)
Because the mass of the ball is 25g or 0.025 kg, the force that the road exerts on the ball is
F = (0.025 kg)*(-2.5 m/s²) = -0.0625 N
The force is negative because it acts in opposition to the motion of the ball.
Answer: 0.0625 N to the left.
Part (b)
The direction of the force exerted by the road is to the left because it opposes the motion of the ball to the right.
Part (c)
The unit force is 1 N (Newton), with the dimension (kg-m)/s².
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Answer:
A. The work done during the process is W = 0
B. The value of heat transfer during the process Q = 442.83
Explanation:
Given Data
Initial pressure = 100 k pa
Initial temperature = 25 degree Celsius = 298 Kelvin
Final pressure = 300 k pa
Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.
⇒ =
------------- (1)
Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.
⇒ P ∝ T
⇒ =
⇒ Put all the values in the above formula we get the final temperature
⇒ =
× 298
⇒ = 894 Kelvin
(A). Work done during the process is given by W = P × ()
From equation (1), =
so work done W = P × 0 = 0
⇒ W = 0
Therefore the work done during the process is zero.
Heat transfer during the process is given by the formula Q = m (
-
)
Where m = mass of the gas = 1 kg
= specific heat at constant volume of nitrogen = 0.743
Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )
⇒ Q = 442.83
Therefore the value of heat transfer during the process Q = 442.83